# HSC Extension 1 and 2 Mathematics/3-Unit/HSC/Primitive of sin and cos squared

The integration of these functions can be performed by the substitution of $\sin^2\alpha$ or $\cos^2\alpha$ by a function in terms of $\cos2\alpha$ respectively.

## Deriving the substituted functions

By the double-angle formula, $\left.\cos 2\alpha = 2\cos^2 \alpha - 1\right.$. Rearranging to make $\cos^2\alpha$ the subject, $\cos^2\alpha = \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha$.

We know by the Pythagorean identity $\left.\sin^2\theta + \cos^2\theta = 1\right.$ that $\left.\cos^2\alpha = 1 - \sin^2\alpha\right.$, so $1 - \sin^2\alpha = \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha$. Rearranging, $\sin^2\alpha = \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha$.

## Finding the primitive

By substituting $\sin^2\alpha\,$ for $\tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha$, $\int \sin^2 \alpha d\alpha = \int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha d\alpha$. Similarly for $\cos^2\alpha \,$,

$\int \cos^2 \alpha \, d\alpha = \int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha$

These can be integrated by using the primitives of cosine and sine

$\int \cos \alpha d\alpha = \sin \alpha + c$

$\int \sin \alpha d\alpha = - \cos \alpha + c$

For cos2 α

$\int \cos^2 \alpha \, d\alpha = \int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha$

$\int \tfrac{1}{2} + \tfrac{1}{2}\cos2\alpha \, d\alpha = \tfrac{1}{2} \alpha + \tfrac{1}{4} \sin2\alpha + c$

$\int \cos^2 \alpha \, d\alpha = \tfrac{1}{2} \alpha + \tfrac{1}{4} \sin2\alpha + c$

For sin2 α

$\int \sin^2 \alpha \, d\alpha = \int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha \, d\alpha$

$\int \tfrac{1}{2} - \tfrac{1}{2}\cos2\alpha \, d\alpha = \tfrac{1}{2} \alpha - \tfrac{1}{4} \sin2\alpha + c$

$\int \sin^2 \alpha \, d\alpha = \tfrac{1}{2} \alpha - \tfrac{1}{4} \sin2\alpha + c$